Cho \(x=2018!.\) Tính \(A=\frac{1}{{{\log }_{{{2}^{2018}}}}x}+\frac{1}{{{\log }_{{{3}^{2018}}}}x}+\,...\,+\frac{1}{{{\log }_{{{2017}^{2018}}}}x}+\frac{1}{{{\log }_{{{2018}^{2018}}}...

* Hướng dẫn giải

Ta có \(A=\frac{2018}{{{\log }_{2}}x}+\frac{2018}{{{\log }_{3}}x}+\,...\,+\frac{2018}{{{\log }_{2017}}x}+\frac{2018}{{{\log }_{2018}}x}=2018\left( {{\log }_{x}}2+{{\log }_{x}}3+\,...\,+{{\log }_{x}}2017+{{\log }_{x}}2018 \right).\)

\(=2018\left( {{\log }_{x}}1+{{\log }_{x}}2+\,...\,+{{\log }_{x}}2017+{{\log }_{x}}2018 \right)=2018.{{\log }_{x}}\left( 1.2.3...2017.2018 \right)=2018.{{\log }_{x}}x=2018.\)

Chọn A.